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3 years ago

Circle which 3 sides would make a triangle

Mathematics

1 answer:

Serhud [2]3 years ago

3 0

All of them because they all have 3 sides

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PLEASE HELP ASAP !!!

BartSMP [9]

You know that a straight line has an angle of 180 degrees;

48+x+x=180 <-- since they are all on the same time
48+2x=180
2x=180-48
2x=132
x=132/2
x=66

Therefore, the x value is 66 degree.

Hope I helped :)

8 0

3 years ago

Can someone check whether its correct or no? this is supposed to be the steps in integration by parts

Gwar [14]

Answer:

$\displaystyle - \int \dfrac{\sin(2x)}{e^{2x}}\: \text{d}x=\dfrac{\sin(2x)}{4e^{2x}}+\dfrac{\cos(2x)}{4e^{2x}}+\text{C}$

Step-by-step explanation:

$\boxed{\begin{minipage}{5 cm}\underline{Integration by parts} \\\\$\displaystyle \int u \dfrac{\text{d}v}{\text{d}x}\:\text{d}x=uv-\int v\: \dfrac{\text{d}u}{\text{d}x}\:\text{d}x$ \\ \end{minipage}}$

Given integral:

$\displaystyle -\int \dfrac{\sin(2x)}{e^{2x}}\:\text{d}x$

$\textsf{Rewrite }\dfrac{1}{e^{2x}} \textsf{ as }e^{-2x} \textsf{ and bring the negative inside the integral}:$

$\implies \displaystyle \int -e^{-2x}\sin(2x)\:\text{d}x$

Using <u>integration by parts</u>:

$\textsf{Let }\:u=\sin (2x) \implies \dfrac{\text{d}u}{\text{d}x}=2 \cos (2x)$

$\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=-e^{-2x} \implies v=\dfrac{1}{2}e^{-2x}$

Therefore:

$\begin{aligned}\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x & =\dfrac{1}{2}e^{-2x}\sin (2x)- \int \dfrac{1}{2}e^{-2x} \cdot 2 \cos (2x)\:\text{d}x\\\\& =\dfrac{1}{2}e^{-2x}\sin (2x)- \int e^{-2x} \cos (2x)\:\text{d}x\end{aligned}$

$\displaystyle \textsf{For }\:-\int e^{-2x} \cos (2x)\:\text{d}x \quad \textsf{integrate by parts}:$

$\textsf{Let }\:u=\cos(2x) \implies \dfrac{\text{d}u}{\text{d}x}=-2 \sin(2x)$

$\textsf{Let }\:\dfrac{\text{d}v}{\text{d}x}=-e^{-2x} \implies v=\dfrac{1}{2}e^{-2x}$

$\begin{aligned}\implies \displaystyle -\int e^{-2x}\cos(2x)\:\text{d}x & =\dfrac{1}{2}e^{-2x}\cos(2x)- \int \dfrac{1}{2}e^{-2x} \cdot -2 \sin(2x)\:\text{d}x\\\\& =\dfrac{1}{2}e^{-2x}\cos(2x)+ \int e^{-2x} \sin(2x)\:\text{d}x\end{aligned}$

Therefore:

$\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{2}e^{-2x}\sin (2x) +\dfrac{1}{2}e^{-2x}\cos(2x)+ \int e^{-2x} \sin(2x)\:\text{d}x$

$\textsf{Subtract }\: \displaystyle \int e^{-2x}\sin(2x)\:\text{d}x \quad \textsf{from both sides and add the constant C}:$

$\implies \displaystyle -2\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{2}e^{-2x}\sin (2x) +\dfrac{1}{2}e^{-2x}\cos(2x)+\text{C}$

Divide both sides by 2:

$\implies \displaystyle -\int e^{-2x}\sin(2x)\:\text{d}x =\dfrac{1}{4}e^{-2x}\sin (2x) +\dfrac{1}{4}e^{-2x}\cos(2x)+\text{C}$

Rewrite in the same format as the given integral:

$\displaystyle \implies - \int \dfrac{\sin(2x)}{e^{2x}}\: \text{d}x=\dfrac{\sin(2x)}{4e^{2x}}+\dfrac{\cos(2x)}{4e^{2x}}+\text{C}$

5 0

2 years ago

Please show working out thanks

IgorLugansk [536]

Answer:

10m x 15m

Step-by-step explanation:

You are given some information.

1. The area of the garden: A₁ = 150m²

2. The area of the path: A₂ = 186m²

3. The width of the path: 3m

If the garden has width w and length l, the area of the garden is:

(1) A₁ = l * w

The area of the path is given by:

(2) A₂ = 3l + 3l + 3w + 3w + 4*3*3 = 6l + 6w + 36

Multiplying (2) with l gives:

(3) A₂l = 6l² + 6lw + 36l

Replacing l*w in (3) with A₁ from (1):

(4) A₂l = 6l² + 6A₁ + 36l

Combining:

(5) 6l² + (36 - A₂)l +6A₁ = 0

Simplifying:

(6) l² - 25l + 150 = 0

This equation can be factored:

(7) (l - 10)*(l - 15) = 0

Solving for l we get 2 solutions:

l₁ = 10, l₂ = 15

Using (1) to find w:

w₁ = 15, w₂ = 10

The two solutions are equivalent. The garden has dimensions 10m and 15m.

3 0

3 years ago

If two 6-sided dice are rolled, what is the<br> probability of rolling double 2s?

Komok [63]

16.7%

There are 6 ways we can roll doubles out of a possible 36 rolls (6 x 6), for a probability of 6/36, or 1/6, on any roll of two fair dice. So you have a 16.7% probability of rolling doubles with 2 fair six-sided dice.

3 0

3 years ago

Read 2 more answers

the scores on your first two history tests were 82 and 95. by how much points did you improve on your second test?

nikklg [1K]

Improved by 13 on your second test.

5 0

3 years ago